576 CHAPTER 29. FIRST ORDER SCALAR ODE

49. Solve: y′+3cos(t)y = 4(cos t)e−3sin t , y(0) = 1.

50. Solve: y′+ tan(t)y = cos(t) ,y(0) =−2.

51. Solve: x2dy+(4x2− xy+3y2

)dx = 0, y(2) =−2.

52. Solve:( 7

2 y−2x)

dx+(x− 9

4 y)

dy = 0 which contains the point (x,y) = (1,2) .

53. Solve: x2dy+(3x2− xy+2y2

)dx = 0, y(2) =−3.

54. Solve:(x3−6x2y− y3

)dx +

(6x3 + xy2

)dy = 0, y(2) = −3. Graph the integral

curve.

55. Solve: (2y−3x)dx+(2x− 4

3 y)

dy= 0 which contains the point (x,y)= (1,2) . Graphthe integral curve.

56. Solve: y′+5cos(3t)y = 2e−(5/3)sin3t cos3t, y(0) = 2.

57. Solve: x2dy+(5x2− xy+5y2

)dx = 0, y(−2) =−2.

58. Solve:(3x+ 19

4 y)

dx+(−4x− 9

4 y)

dy = 0 which contains the point (x,y) = (1,2) .

59. Solve:(x3−3x2y− y3

)dx+

(3x3 + xy2

)dy = 0, y(3) =−1.

60. Solve: (y)dx+(x+4y)dy = 0 which contains the point (x,y) = (1,2) .

61. Solve: 5(t6)

y+ y′ =−5t6et7, y(1) = 1.

62. Solve: x2dy+(6x2− xy+5y2

)dx = 0, y(3) = 3.

63. Find the solutions to the equation y′+ y(3cos t) = 3(cos t)e−3sin t .

64. Solve: (y−2x)dx+( 9

2 y− x)

dy = 0 which contains the point (x,y) = (1,2) .

65. Solve: x2dy+(2x2− xy+ y2

)dx = 0, y(2) =−1.

66. Find the solutions to the equation y′+2ty = tet2.

67. Solve:( 7

3 y−2x)

dx+(x− 4

3 y)

dy = 0 which contains the point (x,y) = (1,2) .

68. Solve: y′+ tan(2t)y = cos2t,y(0) = 2.

69. Find the general solution to the equation

y′+(4x3 + x2 +3x

)y = exp

(−x4− 1

3x3− 3

2x2)

ln(x+1)

70. Show that the following initial value problem fails to have a unique solution.

y′ = y1/(2n+1),y(0) = 0,n a positive integer.

57649.50.S1.52.53.54.55.56.57.58.59.60.61.62.63.64.65.66.67.68.69.70.CHAPTER 29. FIRST ORDER SCALAR ODESolve: y’ +3cos (t) y = 4 (cost) e375", y (0) = 1.Solve: y’ + tan (t) y = cos(t),y(0) = —2.Solve: x7dy+ (4x? —xy +3y) dx =0, y(2) = —2.Solve: (fy — 2x) dx+ (x— $y) dy = 0 which contains the point (x,y) = (1,2).)Solve: x*dy + (3x? — xy + 2y*) dx =0, y(2) = —3.Solve: (x3 — 6x°y—y3) dx + (6x7 cea dy = 0, y(2) = —3. Graph the integralcurve.Solve: (2y — 3x) dx+ (2x — zy) dy =0 which contains the point (x,y) = (1,2). Graphthe integral curve.Solve: y’ +5cos (3t) y = 2e~ 9/3) in3! cos 3t, y(0) = 2.Solve: x7dy + (5x? —xy+ Sy’) dx = 0, y(—2) = —2.Solve: (3x+ 2y) dx+ (—4x— fy) dy =0 which contains the point (x,y) = (1,2).Solve: (x3 — 3x°y — y3) dx + (3x3 +.xy”) dy =0, y(3) = -1.Solve: (y) dx+ (x + 4y) dy = 0 which contains the point (x,y) = (1,2).Solve: 5 (1°) y+y! = — 51%", y(1) =1.Solve: x*dy+ (6x2 —xy+5y’) dx = 0, y(3) =3.Find the solutions to the equation y’ + y(3cost) = 3 (cost) e735",Solve: (y—2x)dx+ (3y —x) dy = 0 which contains the point (x,y) = (1,2).Solve: x°dy + (2x? —xy+y?) dx =0, y(2) =-1.Find the solutions to the equation y’ + 2ty = te.Solve: (fy — 2x) dx+ (x— $y) dy = 0 which contains the point (x,y) = (1,2).Solve: y’ + tan (2t) y = cos 2t,y (0) = 2.Find the general solution to the equation1 3y+ (4x3 +x° + 3x) y = exp (-* — se — >) In(x+1)Show that the following initial value problem fails to have a unique solution.y’ =y!/2"+) (0) = 0,1 positive integer.